Minimum Deletions & Insertions to Transform a String into another

Problem Statement

• Example 1

Input: s1 = "abc"
       s2 = "fbc"
Output: 1 deletion and 1 insertion.
Explanation: We need to delete {'a'} and insert {'f'} to s1 to transform it into s2.

• Example 2

Input: s1 = "abdca"
       s2 = "cbda"
Output: 2 deletions and 1 insertion.
Explanation: We need to delete {'a', 'c'} and insert {'c'} to s1 to transform it into s2.

• Example 3

Input: s1 = "passport"
       s2 = "ppsspt"
Output: 3 deletions and 1 insertion
Explanation: We need to delete {'a', 'o', 'r'} and insert {'p'} to s1 to transform it into s2.

Solution

This problem can easily be converted to the Longest Common Subsequence (LCS). If we can find the LCS of the two input strings, we can easily find how many characters we need to insert and delete from s1. Here is how we can do this:

• Let’s assume len1 is the length of s1 and len2 is the length of s2.

• Now let’s assume c1 is the length of LCS of the two strings s1 and s2.

• To transform s1 into s2, we need to delete everything from s1 which is not part of LCS, so minimum deletions we need to perform from s1 => len1 - c1

• Similarly, we need to insert everything in s1 which is present in s2 but not part of LCS, so minimum insertions we need to perform in s1 => len2 - c1

Let’s jump directly to the bottom-up dynamic programming solution:

Bottom-up Dynamic Programming

Here is the code for our bottom-up dynamic programming approach:

def find_MDI(s1, s2):
    c1 = find_LCS_length(s1, s2)
    print("Minimum deletions needed: " + str(len(s1) - c1))
    print("Minimum insertions needed: " + str(len(s2) - c1))


def find_LCS_length(s1,  s2):
    n1, n2 = len(s1), len(s2)
    dp = [[0 for _ in range(n2+1)] for _ in range(n1+1)]
    maxLength = 0
    for i in range(1, n1+1):
        for j in range(1, n2+1):
            if s1[i - 1] == s2[j - 1]:
                dp[i][j] = 1 + dp[i - 1][j - 1]
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

            maxLength = max(maxLength, dp[i][j])

    return maxLength


def main():
    find_MDI("abc", "fbc")
    find_MDI("abdca", "cbda")
    find_MDI("passport", "ppsspt")


main()

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